We set \(m=\frac{l-1}{k}\) . Note that \(x\vert _{[0, (l-1)T]}\)  is the \(m\) -th iteration of \(x\vert _{[0, kT]}\) . By (REF ), (REF ), (REF ) and Lemma 4.1 in [1]}, we obtain \(m=1\)  and then \(k=l-1\) . From the above process (3.12)-(3.13) and (3.15)-(3.17), we obtain \(k=l-1\)  provided \(e(P^{-1}\gamma (T))=2n\) , and
\(\begin{split}i^{P}(\gamma ,1)&= \dim \ker _{\mathbb {R}}(P-I);\\i^{P}(\gamma ,l)&=i^{P}(\gamma ,k+1)= \dim \ker _{\mathbb {R}}(P-I)+1,\\\nu ^{P^{k}}(\gamma , k)&=\nu ^{P}(\gamma , 1)=1.\end{split}\) 
